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2 years ago

Lance armstrong biking over 40 miles in one day would be considered muscular strength. true or false

Physics

1 answer:

Sonja [21]2 years ago

4 0

True.

Reason:

Muscular strength is the maximum force exerted by a muscle against some form of resistance.

In this example, lance Armstrong is continuously working against the gravitational force. Thus, it can be considered as muscular strength.

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Identify the labeled parts in the figure.

Hitman42 [59]

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

5 0

2 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,